Which tactic is more justified?
1) when the offset is allowed from zero point to the right the player starts betting that the next move the point will return to zero (shifts to the left). when the point returns to zero, a positive result is achieved and the cycle repeats. So say we put on the mathematical expectation.
2) when the offset is allowed from the zero point to the right, the player starts betting that the next point will not return to zero (it will move further to the right). When the point is moved to a certain distance, the game ends with a positive result. When the point returns to zero, the cycle is repeated. So say we set the deviation.
quote:
In the symmetric case, the time until the N-th return to zero grows as N2, and the average number of returns per 2n steps increases as √n. This leads to an unexpected conclusion: in symmetric C. b. The intervals between successive returns to zero become strikingly long.
what tactic is more justified from the point of view of probability theory?
Quote: evsa game with a coin can be considered as a one-dimensional random walk. When the eagle falls out, the point is shifted to the right, tails to the left.
1) when the offset is allowed from zero point to the right the player starts betting that the next move the point will return to zero (shifts to the left). when the point returns to zero, a positive result is achieved and the cycle repeats. So say we put on the mathematical expectation.
2) when the offset is allowed from the zero point to the right, the player starts betting that the next point will not return to zero (it will move further to the right). When the point is moved to a certain distance, the game ends with a positive result. When the point returns to zero, the cycle is repeated. So say we set the deviation.
quote:
In the symmetric case, the time until the N-th return to zero grows as N2, and the average number of returns per 2n steps increases as √n. This leads to an unexpected conclusion: in symmetric C. b. The intervals between successive returns to zero become strikingly long.
what tactic is more justified from the point of view of probability theory?
Neither tactic works. No matter how many points left or right at any time, the next coin flip is 50/50. A prior flip or flips makes no difference.
ZCore13
Quote: evsI know it, but from the quote you can see that the point will be removed from zero!
Doesn't matter. It could be 1 point to the left, 2 points to the right, 50 points any way you want. The next coin flip is 50/50 whether you think one side is due one way or another.
ZCore13
For example:
10 coin tosses, average will be 5, getting 4 or 6 is not unusual, but <=2 or >=8 (-3 and +3, respectively) is kind of unlikely (10.9% to be precise).
But after 1,000 coin tosses, average will be 500, getting 501 or 499 is also not unusual, but 503 or 497 (also +3 and -3, respectively) is NOT unlikely at all. You have to go much further to be in the "outlier" area of ~ 10%. That would be <=473 or >=527 (a 9.37% chance). You can't even go 27 points away if you only do 10 tosses!
I tried it two different ways in Excel and got an error, which is pretty bad given that it can deal with 10^-307 to 10^308, a huge range.
The answer is (1/2)^10000 * (10000!)/( (5000!) * (5000!) )
which can also be written as
(1/2)^10000 * (10000 * 9999 * 9998 * 9997 * ... * 5001) / (5000 * 4999 * 4998 * ... * 3 * 2 * 1)
It will happen about 1 in 125 trials, so it's not a remote occurrence at all. Various ways to calculate, the easiest being 1 / (5000 * pi)^.5. That is just a simplified version of Sterling's approximation.Quote: KevinAAApproximately zero.
Or use the excel function 1 / NORM.DIST(5000,5000,50,0) = 125. 50 being the standard deviation (1/2 * 1/2 * 10,000)^.5
If you find yourself without a calculator, you can even use a z-score table. 1/100 deviations from the mean is a value of .004, multiplied by 2 = .008 = 1/125, since we are looking for 1/50th of a deviation.
Quote: KevinAAApproximately zero.
I tried it two different ways in Excel and got an error, which is pretty bad given that it can deal with 10^-307 to 10^308, a huge range.
The answer is (1/2)^10000 * (10000!)/( (5000!) * (5000!) )
which can also be written as
(1/2)^10000 * (10000 * 9999 * 9998 * 9997 * ... * 5001) / (5000 * 4999 * 4998 * ... * 3 * 2 * 1)
Using Excel and logarithms, note that:
(10000 * 9999 * 9998 * 9997 * ... * 5001) / (5000 * 4999 * 4998 * ... * 3 * 2 * 1) = 10^(log 5001 - log 1 + log 5002 - log 2 + ... + log 10000 - log 5000)
and (1/2)^10000 = 2^(-10000) = 10^(-10000 log 2)
The product is 10^(log 5001 - log 1 + log 5002 - log 2 + ... + log 10000 - log 5000 - 10000 log 2), which, according to LibreOffice Calc, is about 1 / 125.335.
In my Excel 2007 it is simple function useQuote: KevinAAApproximately zero.
I tried it two different ways in Excel and got an error,
=BINOMDIST(5000,10000,0.5,FALSE)
0.0079786461
about 1 in 125.33
I think reading once that over 1000 trials the function uses the normal distribution
in R from this calculator using section 2r.
https://sites.google.com/view/krapstuff/coins
> data(10000, 1/2, 5000, 5000)
successes probability
[1,] 5000 0.007978646
they both seem to agree
Sally
here is some dataQuote: KevinAABut after 1,000 coin tosses, average will be 500, getting 501 or 499 is also not unusual, but 503 or 497 (also +3 and -3, respectively) is NOT unlikely at all.
> data(1000, 1/2, 497, 503)
successes probability cumulative (for range)
[1,] 497 0.0247754729 0.0247754729
[2,] 498 0.0250242226 0.0497996955
[3,] 499 0.0251746688 0.0749743643
[4,] 500 0.0252250182 0.100199383
[5,] 501 0.0251746688 0.125374051
[6,] 502 0.0250242226 0.150398274
[7,] 503 0.0247754729 0.175173747
most really have no clue on distribution probabilities (or for that matter the Law of Large Numbers)
except the belief in the 'law of averages' and that is the false belief that as more fair coin tosses happen one is more likely to get exactly 50% of those tosses as success. (500 out of 1000 for example, not just closer to 50%)
this was so true when I was in math classes (belief in the law of averages)
of course this is NOT true, but more do believe against the truth than with it
just the way math is perceived by all
I mean, IF all math was so easy
all would be doing it and maybe no one would make any money from it
Sally
1 /125 fits my definition of pretty slim.
Many players mis-apply the concept of 'the law of large numbers' and this may be tripping you up too. It is true that an anomaly like 67% heads tossed in small numbers will almost certainly disappear in large numbers, but the way this happens is not by the coins developing a 'will to correct' - being inanimate objects. The anomaly disappears because the larger set of numbers will cause that anomaly to become insignificant.Quote: evsI know it, but from the quote you can see that the point will be removed from zero!
but most do not believe that.Quote: billryanMy point was that while long term results should be around 50-50, the chances of them being exactly 50-50 were pretty slim.
1 /125 fits my definition of pretty slim.
the law of averages says so, and most believe that.
it is easy to understand that and most do
say 10 fair coin flips
most believe that 5 heads and 5 tails will happen about 50% of the time
it is what 'should happen'
they come away with this idea in math class from school (I did)
> data(10, 1/2)
successes probability cumulative (for range)
[1,] 0 0.0009765625 0.0009765625
[2,] 1 0.009765625 0.0107421875
[3,] 2 0.0439453125 0.0546875
[4,] 3 0.1171875 0.171875
[5,] 4 0.205078125 0.376953125
[6,] 5 0.24609375 0.623046875
[7,] 6 0.205078125 0.828125
[8,] 7 0.1171875 0.9453125
[9,] 8 0.0439453125 0.989257813
[10,] 9 0.009765625 0.999023438
[11,] 10 0.0009765625 1
and not one person can tell with an easy proof that they are wrong
it is funny that way
how people believe what they want to about math
and some get real mad about it too
Sally
